Checking command line argument in PHP

Question

I want to pass two different arguments to my script and based on the sent argument, I need my script does something.But I don't know how to define my conditional statement.

To be more precise, I want my script does searching when I pass "search" argument and alternatively showing the result when I pass "show" argument.

Here is my code:

if ($argc > 1) {
  if ($argv[0] == 'show') {
    for ($i = 0; $i <= $argv[2]; $i++) {
      //do something
    }
  }
  elseif($argv[0] == 'search') {
    //do something
  }
} else {
  echo "no argument passed\n";
}

The "IF" statement is not checking my passing argument whether it is "search" or "show"

Answer 1

$argv[0] is the name of the script, that's why your code doesn't work.

If I have a file script.php:

<?php
if ($argc > 1) {
  if ($argv[1] == 'show') {
    for ($i = 0; $i <= $argv[2]; $i++) {
      print "show passed\n";
    }
  }
  elseif($argv[1] == 'search') {
    print "search passed";
  }
} else {
  echo "no argument passed\n";
}

Testing gives:

$php script.php

no argument passed

$php script.php search

search passed

$php script.php show 2

show passed
show passed
show passed